Cambridge A Level Chemistry 9701 — 2013 May/June Paper 4 · Variant 2

READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fl uid. DO NOT WRITE IN ANY BARCODES. Section A Answer all questions. Section B Answer all questions. You may lose marks if you do not show your working or if you do not use appropriate units. A Data Booklet is provided. Electronic calculators may be used. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. CHEMISTRY 9701/42 Paper 4 Structured Questions May/June 2013 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certifi cate of Education Advanced Level This document consists of 16 printed pages and 4 blank pages. [Turn over IB13 06_9701_42/5RP © UCLES 2013 *3126331034* For Examiner’s Use 1 2 3 4 5 6 7 8 Total

2 9701/42/M/J/13 © UCLES 2013 For Examiner’s Use Section A Answer all the questions in the spaces provided. 1 A bromoalkane, R–Br, is hydrolysed by aqueous sodium hydroxide. (a) (i) Write a balanced equation for this reaction. … (ii) What type of reaction is this? … [2] (b) The concentration of bromoalkane was determined at regular time intervals as the reaction progressed. Two separate experiments were carried out, with different NaOH concentrations. The graph below shows the results of an experiment using [NaOH] = 0.10 mol dm–3. 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010 0 50 100 150 200 250 time / min [R–Br] / mol dm–3 When the experiment was repeated using [NaOH] = 0.15 mol dm–3, the following results were obtained. time / min [R–Br] / mol dm–3 0 0.0100 40 0.0070 80 0.0049 120 0.0034 160 0.0024 200 0.0017 240 0.0012 (i) Plot these data on the axes above, and draw a line of best fi t.

3 9701/42/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (ii) Use one of the graphs to confi rm that the reaction is fi rst order with respect to R–Br. Show all your working, and show clearly any construction lines you draw. (iii) Use the graphs to calculate the order of reaction with respect to NaOH. Show all your working, and show clearly any construction lines you draw on the graphs. (iv) Write the rate equation for this reaction, and calculate the value of the rate constant. rate = [7] (c) Nitric oxide, NO, and bromine vapour react together according to the following equation. 2NO(g) + Br2(g) → 2NOBr(g) ∆H = –23 kJ mol–1 The reaction has an activation energy of +5.4 kJ mol–1 . Use the following axes to sketch a fully-labelled reaction pathway diagram for this reaction. Include all numerical data on your diagram. energy / kJ mol–1 2NO + Br2 extent of the reaction [2] [Total: 11]

4 9701/42/M/J/13 © UCLES 2013 For Examiner’s Use 2 (a) (i) With the aid of a fully-labelled diagram, describe the standard hydrogen electrode. (ii) Use the Data Booklet to calculate the standard cell potential for the reaction between Cr2+ ions and Cr2O7 2– ions in acid solution, and construct a balanced equation for the reaction. = … V equation … (iii) Describe what you would see if a blue solution of Cr2+ ions was added to an acidifi ed solution of Cr2O7 2– ions until reaction was complete. … … [8]

5 9701/42/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (b) A buffer solution is to be made using 1.00 mol dm–3 ethanoic acid, CH3CO2H, and 1.00 mol dm–3 sodium ethanoate, CH3CO2Na. Calculate to the nearest 1 cm3 the volumes of each solution that would be required to make 100 cm3 of a buffer solution with pH 5.50. Clearly show all steps in your working. Ka (CH3CO2H) = 1.79 × 10–5 mol dm–3 volume of 1.00 mol dm–3 CH3CO2H = … cm3 volume of 1.00 mol dm–3 CH3CO2Na = … cm3 [4] (c) Write an equation to show the reaction of this buffer solution with each of the following. (i) added HCl … (ii) added NaOH … [2] (d) Choose one reaction in organic chemistry that is catalysed by an acid, and write the structural formulae of the reactants and products in the boxes below. H+ [3] [Total: 17]

6 9701/42/M/J/13 © UCLES 2013 For Examiner’s Use 3 (a) Describe the reagents and conditions required to form a nitro compound from the following. CH3 methylbenzene … (i) OH phenol … (ii) [3] (b) Draw the structure of the intermediate organic ion formed during the nitration of benzene. [1] (c) In the box over the arrow below, write the reagents needed to convert nitrobenzene into phenylamine. NO2 NH2 [1]

7 9701/42/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (d) Phenylamine can be converted into the organic compounds A and B. (i) Suggest the structural formulae of A and B in the boxes below. (ii) Suggest suitable reagents and conditions for step 1, and write them in the box over the arrow. step 1 NH2 N+ O– Na+ Cl – N Br2(aq) B A [3] (e) When phenylamine is treated with propanoyl chloride a white crystalline compound, C, C9H11NO, is formed. (i) Name the functional group formed in this reaction. … (ii) Calculate the percentage by mass of nitrogen in C. percentage = … % (iii) Draw the structural formula of C. [3] [Total: 11]

8 9701/42/M/J/13 © UCLES 2013 For Examiner’s Use 4 (a) (i) Suggest why transition elements show variable oxidation states in their compounds whereas s-block elements like calcium do not. … … (ii) Calculate the oxidation number of the metal in each of the following ions. VO2 + … CrF6 2– … MnO4 2– … [4] (b) Explain why transition element complexes are often coloured whereas compounds of s-block elements such as calcium and sodium are not. … … … … … … [4] (c) SO2 and MnO4 – react together in acidic solution. (i) Use the Data Booklet to construct a balanced equation for this reaction. … (ii) Describe the colour change you would see when SO2(aq) is added to a sample of acidifi ed KMnO4 until the SO2 is in excess. from … to … [3] (d) Describe the observations you would make when NH3(aq) is added gradually to a solution containing Cu2+ ions, until the NH3 is in an excess. … … … … [3] [Total: 14]

9 9701/42/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 5 Coffee beans contain chlorogenic acid. O O O HO HO OH OH chlorogenic acid OH OH (a) (i) Draw circles around any chiral centres in the above structure. (ii) Write down the molecular formula of chlorogenic acid. … (iii) How many moles of H2(g) will be evolved when 1 mol of chlorogenic acid reacts with an excess of sodium metal? … (iv) How many moles of NaOH(aq) will react with 1 mol of chlorogenic acid under each of the following conditions? in the cold … on heating … [6]

10 9701/42/M/J/13 © UCLES 2013 For Examiner’s Use (b) On heating with dilute aqueous acid, chlorogenic acid produces two compounds, D and E. chlorogenic acid dil. H+(aq) heat O HO HO HO OH + OH OH D E O OH OH conc. H2SO4 heat F G C7H6O3 Br2(aq) in excess (i) What type of reaction is chlorogenic acid undergoing when D and E are formed? … When compound D is heated with concentrated H2SO4, compound F, C7H6O3, is formed. Compound F evolves CO2(g) when treated with Na2CO3(aq), and decolourises Br2(aq), giving a white precipitate. It does not, however, decolourise cold dilute acidifi ed KMnO4. When compound E is treated with an excess of Br2(aq), compound G is produced. (ii) If the test with cold dilute acidifi ed KMnO4 had been positive, which functional group would this have shown to be present in F? … (iii) Name the functional groups in compound F that would react with the following. Na2CO3(aq) … Br2(aq) … (iv) Suggest structures for compounds F and G and draw them in the relevant boxes above.

11 9701/42/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (v) Compound E is one of a pair of stereoisomers. What type of stereoisomerism is shown by compound E? … (vi) Draw the structure of the other stereoisomer in the box below. [8] (c) Calculate the volume of 0.1 mol dm–3 NaOH that is needed to react completely with 0.1 g of compound E. volume = … cm3 [3] [Total: 17]

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13 9701/42/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use Section B Answer all the questions in the spaces provided. 6 There are two important polymerisations that occur within living organisms – protein synthesis and the formation of DNA. (a) Complete the table by placing a tick ( ) in the correct column to indicate in which process each substance could be used. substance protein synthesis formation of DNA cysteine cytosine glutamine guanine [3] (b) DNA consists of a double helical structure. (i) Describe the bonding between the two strands in DNA and state which part of each strand is joined by it. … … (ii) How does the strength of this bonding relate to the mechanism of the replication of DNA? … … [4] (c) Some diseases are caused by changes in the structure of proteins. Explain the genetic basis of these changes. … … … … [3] [Total: 10] 

14 9701/42/M/J/13 © UCLES 2013 For Examiner’s Use 7 The techniques of mass spectrometry and NMR spectroscopy are useful in determining the structures of organic compounds. (a) The three peaks of highest mass in the mass spectrum of organic compound L correspond to masses of 142, 143 and 144. The ratio of the heights of the M : M+1 peaks is 43.3 : 3.35, and the ratio of heights of the M : M+2 peaks is 43.3 : 14.1. (i) Use the data to calculate the number of carbon atoms present in L. (ii) Explain what element is indicated by the M+2 peak. … … Compound L reacts with sodium metal. The NMR spectrum of compound L is given below. 9 8 7 6 5 4 3 2 1 0 δ / ppm 4 2 1 (iii) What does the NMR spectrum tell you about the number of protons in L and their chemical environments? … …

15 9701/42/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (iv) Use the information given and your answers to (i), (ii) and (iii) to deduce a structure for L. Explain how you arrive at your answer. structure of L [7] (b) The molecular formula C3H6 represents the compounds propene and cyclopropane. CH3CH CH2 propene cyclopropane C H H C H H C H H (i) Suggest one difference in the fragmentation patterns of the mass spectra of these compounds. … … (ii) Suggest two differences in the NMR spectra of these compounds. … … … [3] [Total: 10]

16 9701/42/M/J/13 © UCLES 2013 For Examiner’s Use 8 In recent years there has been considerable interest in a range of polymers known as ‘hydrogels’. These polymers are hydrophilic and can absorb large quantities of water. (a) The diagram shows part of the structure of a hydrogel. C O O CH O O CH2 CH2 CH C O O C C CH2 C HO CH2 CH2 CH2 OH CH2 CH CH O O CH2 The hydrogel is formed from chains of one polymer which are cross-linked using another molecule. (i) Draw the structure of the monomer used in the polymer chains. (ii) State the type of polymerisation used to form these chains. … … (iii) Draw the structure of the molecule used to cross-link the polymer chains.

17 9701/42/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (iv) During the cross-linking, a small molecule is formed as a by-product. Identify this molecule. … [5] (b) Once a hydrogel has absorbed water, it can be dried and re-used many times. Explain why this is possible, referring to the structure on the opposite page. … … … [2] (c) Not every available side chain in the polymer is cross-linked, and the amount of cross-linking affects the properties of the hydrogel. (i) The amount of cross-linking has little effect on the ability of the gel to absorb water. Suggest why this is the case. … … … (ii) Suggest one property of the hydrogel that will change if more cross-linking takes place. Explain how the increased cross-linking brings about this change. … … … [3] [Total: 10]

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20 Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. 9701/42/M/J/13 © UCLES 2013 BLANK PAGE

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2013 series 9701 CHEMISTRY 9701/42 Paper 4 (A2 Structured Questions), maximum raw mark 100 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 42 © Cambridge International Examinations 2013 1 (a) (i) RBr + OH– → ROH+ Br– [1] (ii) nucleophilic substitution [1] [2] (b) (i) plotting of all points (plotted to within ½ small square) [1] good line of best fit [1] (ii) t ½ = 118 min or 79 min (± 5 min) or construction lines for two half-lives and mention that half-life is constant or calculate the ratio of two rates at two different concentrations [1] (iii) either ratio of initial rates (slopes) or ratio of t ½ or ratio of times for [RBr] to fall to the same level: all should be = 1.5 [1] therefore reaction is first order w.r.t. [OH–] [1] (iv) rate = k [RBr] [OH–] [1] initial rate = 0.01 / 185 = 5.4 × 10–5 (mol dm–3 min–1) [1] k = 5.4 x 10-5 / (0.01 × 0.1) = 0.054 (mol–1 dm3 min–1) [1] [8 max 7]

Page 3 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 42 © Cambridge International Examinations 2013 (c) extent of reaction energy/kJ moli1 (2NO + Br2) 2NOBr +5.4 -23 four marking points: one activation "hump" 2NOBr (not just NOBr) ∆H labelled correctly (arrow down, or double headed, or just a line) Ea labelled correctly (arrow up, or double headed, or just a line) all four points [2] three or two points [1] [2] [Total: 11]

Page 4 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 42 © Cambridge International Examinations 2013 2 (a) (i) hydrogen gas at 1 atm Pt H+/HCl at 1 mol dm-3 and 298K H2(g) going in (i.e. not being produced) [1] platinum electrode in contact with solution, with H2 bubbling over it [1] H+ or HCl or H2SO4 [1] solution at 1 mol dm–3(or 0.5 M if H2SO4) and T=298 K, p=1 atm [1] (ii) Eo = 1.33 – (-0.41) = 1.74 V [1] Cr2O7 2- + 14H+ + 6Cr2+ → 8Cr3+ + 7H2O [1] (iii) Colour would change from orange [1] to green [1] [8] (b) there are two ways of calculating the ratio: pKa = –log10(Ka) = –log10(1.79 x 10–5) = 4.747 (4.75) or [H+] = 10-5.5 = 3.16 x 10-6 [1] log10([B] / [A]) = pH – pKa = 0.753 (0.75) or [salt] / [acid] = Ka / [H+] [1] ∴ [B] / [A] = 100.753 = 5.66 or = 1.79 x 10–5 / 3.16 x 10–6 = 5.66 (or [A] / [B] = 0.177) [1] (correct ratio = [3] marks) since B + A = 100,∴ (100–A) / A = 5.66 ⇒ vol of acid = 15 cm3 vol of salt = 85 cm3 [1] [4]

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 42 © Cambridge International Examinations 2013 (c) (i) CH3CO2Na + HCl → CH3CO2H + NaCl [1] (ii) CH3CO2H + NaOH → CH3CO2Na + H2O [1] [2] (d) e.g. hydrolysis of esters RCO2R' (+ H2O) → RCO2H + R'OH or its reverse or hydrolysis of amides: RCONH2 (+ H3O+) → RCO2H + NH4 + hydrolysis of nitriles: RCN (+ H3O+ + H2O) → RCO2H + NH4 + nitration of benzene (or any arene): C6H6 + HNO3 → C6H5NO2 (+ H2O) dehydration of alcohols, e.g. : CH3CH(OH)CH3 → CH3CH=CH2 + H2O (or the reverse) halogenation of ketones, e.g. : CH3COCH3 + X2 → CH3COCH2X (+ HX) [3] [Total: 17] 3 (a) (i) HNO3 + H2SO4 [1] conc (both acids) and 30oC < T < 60oC or warm [1] (ii) dilute HNO3 or HNO3(aq) and room temp. (allow T ≤ 30oC) [1] [3] (b) (allow intermediate from methylbenzene) H NO2 [1] [1] (c) Sn/tin (or SnCl2, Fe) + HCl (NOT H2SO4 or H+, Zn, or LiAlH4.) [1] [1]

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 42 © Cambridge International Examinations 2013 (d) (i) N N OH NH2 Br Br Br A B or -ONa or -O- or NH3 + NOT -NaO [1] + [1] (ii) NaNO2 + HCl or H2SO4 or H+ or HNO2 [1] T ≤ 10oC [1] [4 max 3] (e) (i) amide [1] (ii) Mr = 108+11+14+16 = 149 %N = (14 x 100)/149 = 9.4% [1] (iii) NHCOC2H5 [1] [3] [Total: 11] 4. (a) (i) Many electrons of similar energy in a valence-shell orbital or successive ionisation energies rise steadily (no big jumps) or ability to form bonds with ligands can stabilise very low or very high oxidation states or 4s + 3d orbitals/shells/energy levels have similar / same energies [1] (ii) VO2 +: +5 CrF6 2–: +4 MnO4 2–: +6 [3 × 1] [4]

Page 7 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 42 © Cambridge International Examinations 2013 (b) • (colour due to) absorption of light/photons/frequencies/wavelengths or colour seen is complement of colour absorbed. • d-orbitals/d-subshell split (by ligand field) • (when photon is absorbed), electron is promoted or moves (from lower) to higher (d–)orbital • energy difference/gap or ∆E or splitting corresponds to photon/frequency/wavelength in visible region • in s-block elements the energy gap is too large (to be able to absorb visible light) [any four 4 × 1] [4] (c) (i) 2MnO4 – + 2H2O + 5SO2 → 2Mn2+ + 5SO4 2– + 4H+ [1] (ii) solution will go from purple [1] to colourless [1] [3] (d) (pale) blue solution [1] gives a (pale) blue ppt. [1] which re-dissolves, or forms a solution, which is dark/deep blue or purple [1] [3] [Total: 14]

Page 8 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 42 © Cambridge International Examinations 2013 5 (a) (i) O O OH O HO OH HO OH OH * * * * two or three centres correctly identified [1] four centres correctly identified [2] (ii) C16H18O9 [1] (iii) 3 moles of H2 [1] (iv) in cold: 3 moles of NaOH [1] on heating: 4 moles of NaOH [1] [6] (b) (i) hydrolysis [1] (ii) alkene or C=C [1] (iii) with Na2CO3(aq): carboxylic acid [1] with Br2(aq): phenol [1] (iv) HO2C OH F CO2H OH OH Br(OH) Br Br Br G (ring subst. allow 2 or 3 Br in ring) [1] (OH can be at the 3, 4, or 5 positions, but not the 2 or 6 positions) (OH) (addition to C=C: allow one of the aliphatic Br to be OH, but not both) [1] [1] (v) geometrical or cis-trans or E-Z [1]

Page 9 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 42 © Cambridge International Examinations 2013 (vi) HO2C OH OH skeletal or structural [1] [9 max 8] (c) Mr(E) = 180, so 0.1 g = 1/1800 (5.56 x 10-4) mol [1] 3 mol NaOH react with 1 mol of E, so n(NaOH) = 3/1800 = 1/600 mol = 1.67 × 10-3 mol [1] volume of 0.1M NaOH = 1000/(600 x 0.1) = 16.7 cm3 [1] [3] [Total: 17]

Page 10 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 42 © Cambridge International Examinations 2013 6 (a) substance protein synthesis formation of DNA cysteine ✓ cytosine ✓ glutamine ✓ guanine ✓ [3] [3] (b) (i) Hydrogen bonding [1] Between bases or between A,T, C and G (all four needed) [1] (ii) Bonds are (relatively) weak or easily broken [1] This enables strands to separate or DNA to unzip/unwind/unravel. [1] [4] (c) changes / mutations in DNA • by the addition / insertion /deletion / substitution / replacement of a base • adds / deletes / replaces an amino acid or changes the amino acid sequence • this causes a loss of function or changes the shape / tertiary structure of the protein any three points [3] [3] [Total: 10]

Page 11 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 42 © Cambridge International Examinations 2013 7. (a) (i) 43.3 = 100 3.35 1.1 x n n = 100 x 3.35 = 7.03 = 7 (calculation must be shown) [1] 43.3 x 1.1 (ii) The M and M+2 peaks are in the ratio 3 : 1 hence the halogen is chlorine/Cl [1] (iii) L contains 7 hydrogen atoms or there are 3 types/environments of proton/H [1] (iv) The multiplet with 4 hydrogens or peaks at δ 7.3 suggests a benzene ring The singlet with 2 hydrogens or peak at δ 4.7 suggests a –CH2– group The singlet with 1 hydrogen or peak at δ 2.3 suggests an –OH group or reaction with Na suggests an OH group OH must be an alcohol, not a phenol (due to its δ value) Since L also contains 7 carbon atoms and chlorine, this accounts for 126 of the 142 mass, the remaining atom must be oxygen Thus L is CH2OH Cl (allow the 2-, 3- or 4- isomer) [6] [9 max 7] (b) (i) we expect propene to have a CH3 peak or a peak at m/e 15 or cyclopropane would have fewer peaks [1] (ii) cyclopropane would have 1 peak (ignore splitting) propene would have 2 (or 3, or 4) peaks (ignore splitting) or propene would have peaks in the δ 4.5-6.0 (alkene) region no splitting of cyclopropane peak (any two points) [2] [3] [Total: 10]

Page 12 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9701 42 © Cambridge International Examinations 2013 8 (a) (i) CH2 = CH–CO2H or CH2 = CH–CO2R or CH2 = CH–COCl [2] (ii) addition (polymerisation) [1] (iii) C(CH2OH)4 [1] (iv) water [1] [5] (b) (water is bonded to the polymer by) hydrogen bonding [1] hydrogen bonds are weak or easily broken [1] [2] (c) (i) cross-linking causes no reduction in the number of –OH groups or cross-linking molecules also have –OH groups [1] (ii) property e.g. becomes harder / more rigid / less flexible / stronger / higher melting point. [1] because the chains are more strongly / tightly held [1] [3] [Total: 10]